### Problem:A student goes to the cafeteria to get an iced tea. He puts 3 ice cubes (approximately 20.0 mL each, \( c_{\text{ice}} = 2.100 \, \text{J/g}^\circ\text{C} \)) from the ice machine (at \(-15.0^\circ\text{C}\)) into a well-insulated cup and then fills the cup with 375 mL of tea. The tea is coming from a machine at room temperature, which is \( 22.5^\circ \text{C} \). The student is so immersed in studying chemistry that he forgets about his drink. When he finally goes to take his first drink, he notices that the ice has just melted. At what temperature is his iced tea? Assume the density of ice is that of water. It may be helpful to plot the temperatures on the heating curve below. \( L_{\text{H}_2\text{O}} = 333 \, \text{J/g} \).### Graph Explanation:The graph depicts a heating curve indicating the temperature changes over time for the iced tea. - The y-axis represents temperature.- The x-axis represents time.The curve starts at a lower temperature and shows an initial phase where the temperature remains constant as the ice absorbs heat and begins to melt, a phase where the temperature increases once all the ice has melted, and then a further temperature increase of the liquid.#### Key Points of the Heating Curve:1. **Initial Horizontal Line**: Represents the phase where the ice is absorbing heat but not changing temperature.2. **Diagonal Line Upward**: Indicates the increasing temperature of the liquid after the ice has melted, as it continues to absorb heat.### Conceptual Explanation:The problem involves calculating the final temperature of the iced tea mixture after the ice has completely melted, considering the heat exchanges between the ice and the tea.

Answered: Image /qna-images/answer/b3cf1149-bde6-4a6e-8ad1-c64aa52a6420.jpg

3. A student goes to the cafeteria to get an iced tea. He puts 3 ice cubes (approximately 20.0 mL each, Sice = 2.100 J/g °C) from the ice machine (at - 15.0°C) into a well- insulated cup and then fills the cup with 375 mL of tea. The tea is coming from a machine at room temperature, which is 22.5°C. The student is so immersed in studying chemistry that he forgets about his drink. When he finally goes to take his first drink, he notices that the ice has just melted. At what temperature is his iced tea? Assume the density of ice is that of water. It may be helpful to plot the temperatures on the heating curve below. H₂O = 333 J/g. Temperature Time

3. A student goes to the cafeteria to get an iced tea. He puts 3 ice cubes (approximately 20.0 mL each, Sice = 2.100 J/g °C) from the ice machine (at - 15.0°C) into a well- insulated cup and then fills the cup with 375 mL of tea. The tea is coming from a machine at room temperature, which is 22.5°C. The student is so immersed in studying chemistry that he forgets about his drink. When he finally goes to take his first drink, he notices that the ice has just melted. At what temperature is his iced tea? Assume the density of ice is that of water. It may be helpful to plot the temperatures on the heating curve below. H₂O = 333 J/g. Temperature Time

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

An aluminum can of mass 0.14kg contains water at 17.5°C. Iron pellets with a total mass of 6.3kg at 85.1°C are dropped into the can, and the can and it's contents come to equilibrium af 32.9°C. Specific heat of water = 4190 J/(kg•°C). Specific heat of aluminum = 910 J/(kg•°C). Specific heat of iron = 470 J/(kg•°C). how many kilograms of water are in the can? (3 sig fig)
A 20.0 g block of iron (molar heat capacity 25.1 J/mol･°C) at 36.6 °C is placed into 200.0 g of water initially at 25.0 °C. What is the change in temperature (in °C) of the iron block? (The molar heat capacity of water is 75.38 J/mol･°C).
Chamistry
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('heavy water,' specific heat capacity = 4.211 J/°C･g) at 25.5°C. The final temperature of the platinum and deuterium oxide mixture is 35.1°C. The density of platinum is 21.45 g/cm³ and the density of deuterium oxide is 1.11 g/mL. What is the edge length of the cube of platinum, in centimeters?
A metal alloy (c = 0.550 J/g°C) with a mass of 275g absorbs 650.0 J. If the initial temperature of the alloy is 25.0°C, what is the finall temperature?
20.0 g of ice cubes at 0.0°C are combined with 150. g of liquid water at 70.0°C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings. (Data: specific heat capacity of H20(), c = 4.18 J/g.°C; H 20( s) – H20() A H= 6.02 kJ/mol)Hint: When you equate heat lost to heat gained the heat capacity is cancelled since it si found on both sides of the equation and this simplifies your calculation. O 0.0 O 10.6 O 30.7 O 56.4 O61.8
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C･g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('heavy water,' specific heat capacity = 4.211 J/°C･g) at 25.5°C. The final temperature of the platinum and deuterium oxide mixture is 31.3°C. The density of platinum is 21.45 g/cm³ and the density of deuterium oxide is 1.11 g/mL. What is the edge length of the cube of platinum, in centimeters?
A soft drink (290 mL of mostly water) initially at 18.0 C is cooled by adding 197 g of ice cubes at 0 C. What mass of ice remains when the temperature of the drink reaches equilibrium (water and ice at 0 C)? The enthalpy of fusion of water (i.e., the heat required to melt ice) is 6.007 kJ/mol. The density of the water is 1.0 g/mL. Assume the specific heat capacity of water is constant and equal to 4.184 J/g/C.